Basic proof of units in a ring with identity.

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Homework Statement



Let R be a ring with identity, and a,b are elements in R. If ab is a unit, and neither a nor b is a zero divisor, prove a and b are units.


Homework Equations



If ab is a unit then (ab)c=1=c(ab) for some c in R.


The Attempt at a Solution


Assume both a and b are not zero divisors, and denote the identity element of R as 1.

Since ab is a unit in R, there exists some c such that (ab)c = 1.
(ab)c=1 [itex]\Rightarrow[/itex] (ca)b = 1 [itex]\Rightarrow[/itex] b is a unit.

Similarly, (ab)c= 1 [itex]\Rightarrow[/itex] a(bc) [itex]\Rightarrow[/itex] a is a unit.

Does the above sufficiently prove the claim in the problem statement? I feel like I'm missing something.

Thanks!
 

Answers and Replies

  • #2
Dick
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Homework Statement



Let R be a ring with identity, and a,b are elements in R. If ab is a unit, and neither a nor b is a zero divisor, prove a and b are units.


Homework Equations



If ab is a unit then (ab)c=1=c(ab) for some c in R.


The Attempt at a Solution


Assume both a and b are not zero divisors, and denote the identity element of R as 1.

Since ab is a unit in R, there exists some c such that (ab)c = 1.
(ab)c=1 [itex]\Rightarrow[/itex] (ca)b = 1 [itex]\Rightarrow[/itex] b is a unit.

Similarly, (ab)c= 1 [itex]\Rightarrow[/itex] a(bc) [itex]\Rightarrow[/itex] a is a unit.

Does the above sufficiently prove the claim in the problem statement? I feel like I'm missing something.

Thanks!

Yes, you should feel uncomfortable with that. You didn't use that a and b weren't zero divisors. If (ab)c=1 then you can regroup that to a(bc)=1. That doesn't prove a is a unit. You also need to show (bc)a=1. Don't assume the ring is commutative.
 
  • #3
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Yes, you should feel uncomfortable with that. You didn't use that a and b weren't zero divisors. If (ab)c=1 then you can regroup that to a(bc)=1. That doesn't prove a is a unit. You also need to show (bc)a=1. Don't assume the ring is commutative.

Ah thank you, that was a foolish oversight, they wouldn't stipulate a and b weren't zero divisors if it wasn't necessary...

So if we let a, b and c be elements of R such that ab = ac, then:

ab-ac=0R = a(b-c) = 0R, and
a is not a zero divisor [itex]\Rightarrow[/itex] (b - c) = 0R [itex]\Rightarrow[/itex] (b-c) = 0R [itex]\Rightarrow[/itex] b = c

Does that allow me to claim the following?

Multiplying a(bc) = 1 by a to give a(bc)a = 1a , (1a=a1 by definition) then, a(bc)a = a1 and by cancellation, (bc)a = 1 [itex]\Rightarrow[/itex] a is a unit.
 
  • #4
Dick
Science Advisor
Homework Helper
26,263
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Ah thank you, that was a foolish oversight, they wouldn't stipulate a and b weren't zero divisors if it wasn't necessary...

So if we let a, b and c be elements of R such that ab = ac, then:

ab-ac=0R = a(b-c) = 0R, and
a is not a zero divisor [itex]\Rightarrow[/itex] (b - c) = 0R [itex]\Rightarrow[/itex] (b-c) = 0R [itex]\Rightarrow[/itex] b = c

Does that allow me to claim the following?

Multiplying a(bc) = 1 by a to give a(bc)a = 1a , (1a=a1 by definition) then, a(bc)a = a1 and by cancellation, (bc)a = 1 [itex]\Rightarrow[/itex] a is a unit.

Yes, that's the idea.
 

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